Page 74 - DCAP313_LAB_ON_COMPUTER_GRAPHICS

Lab on Computer Graphics

Notes y is always decremented in each step, i.e. y k+1 = y k – 1.
x k+1 = x k if S is selected, or x k+1 = x k + 1 if SE is selected.
P k = f(x k + ½, y k – 1)

2
2
2
2 2
2
= b (x k + ½) + a (y k – 1) – a b
-
2
= b(x+ x+ 1 4 )+ a(y - 2y+ 1) ab 2
2
2
2
2
k
k
k
k
If P k > 0, select S:
P k+1 = f(x k + ½, y k – 2)
S
= b (x k + ½) + a (y k – 2) – a b
2 2
2
2
2
2
2
2
2
2
2
= b(x+ x+ 1 4 )+ a(y - 4y+ 4) - ab 2
k
k
x
k
S
S
2
Change of P is: DP = P k+1 - P =a (3 - 2y )
S
k
k
k
k
If P k < 0, select SE:
P k+1 = f(x k + 3/2, y k – 2)
SE
2
= b (x k + 3/2) + a (y k – 2) – a b
2
2 2
2
2
2
2
2
-
2
= b(x+ 3x +9/4)+ a(y - 4y+ 4) ab 2
2
k
k
k
k
2
Change of P is DP = P k+1 – P k = 2b (x k + 1) + a (3 – 2y k )
2
SE
SE
SE
k
k
Calculate the changes of DP :
k
If S is selected,
Dp S k+1 = a (5 – 2y k )
2
S
S
S
D P = DP k+1 - DP =2a 2
2
k
k
DP k+1 = 2b (x k + 1) + a (5 – 2y k )
SE
2
2
SE
2
SE
SE
D P = DP k+1 - DP =2a 2
k
k
If SE is selected,
S
DP k+1 = a (5 – 2y k )
2
D P = DP k+1 - DP =2a 2
S
S
S
2
k
k
2
SE
2
DP k+1 = 2b (2x k + 2) – a (5 – 2y k )
D P = DP k+1 - DP =2(a +b )
SE
2
2
2
SE
SE
k
k
Determine the boundary between region I and II:
dy - bx
Set f(x, y) = 0, =
dx a 2 1 x/a 2
2
-
a 2 b 2
When dy/dx = –1, x = x= andy = .
2
a+ b 2 a+ b 2
2
a 2 b 2
At region I, dy/dx > – 1, x < and y > , therefore
2
2
a+ b 2 a+ b 2
Ê 2a 2 ˆ Ê b 2 ˆ
2
SE
2
DP< b Á Á Ë a+ b 2 +3˜ - 2a Á Á Ë a+ b 2 - 1˜ = 2a + 3b .
2
2
˜
k
˜
2
2
¯
¯
Initial values at region II:
a 2 b 2
x 0 = andy =
0
2
2
a+ b 2 a+ b 2
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